V. Thermochemistry
Key focus of this chapter: enthalpy change
This chapter reviews enthalpy changes, internal energy, pressure–volume work, heat transfer, Hess’s law, entropy, and Gibbs free energy.
Thermochemistry: the study of heat absorbed or released during chemical reactions and physical changes.
A. Energy (E)
: ability to do work (W) or transfer heat (Q).
ΔE (change in internal energy) = Efinal -Einitial
- System: the part of the universe selected for study, such as the reacting substances.
- Surroundings: the rest of the universe
(+) energy | (-) energy | |
W | • Work is done on system | • Work is done by system |
Q (heat) Qp = ΔH at constant pressure, with only pressure–volume work. | • Absorbing heat into system from surroundings | • Releasing heat from system to surroundings |
- Question: What is ΔE when a system does 100 J of work and releases 80 J of heat?
ΔE = W + Q
= (-100J) + (-80J)
= -180J
B. Pressure–volume work
1. Pressure
- 1 atm = 760 mmHg (torr)
2. Work (W)
: pressure–volume work is W = −PextΔV at constant external pressure. More generally, mechanical work is force times displacement in the direction of the force.
+W | 
| • Work is done on system • - ΔV (contraction of volume) • External pressure compresses the system |
- W | 
| • Work is done by system • + ΔV (expansion of volume) • The system expands against external pressure |
- Question: What is the work when a gas expands from 8.0 L to 10.0 L against an external pressure of 3.0 atm?
W = −PextΔV (work is done by system)
= – 3 atm × (10.0 – 8.0 ) L
= −6.0 L·atm ≈ −6.1 × 102 J
C. Heat energy, Q
- Calorie: approximately the energy required to raise the temperature of 1 g of water by 1 °C
** 1 cal = 4.184 J
- Specific heat capacity, C: the heat required to raise the temperature of one gram of a substance by one degree Celsius (or one kelvin).
** Specific heat of water is 4.18 J/(g·°C)
- Question: Calculate the heat energy to raise the temperature of 100 g of water from 50 °C to 80 °C.
Q = C·m·ΔT
= 4.18 J/(g·°C) × 100 g × (80-50) °C
= 12,540 J
- Question: Calculate the specific heat of gold when 650 J of heat is used to increase the temperature of 50 g of gold from 25 °C to 125 °C.
D. Laws of thermodynamics
First law | • Law of conservation of energy: the total energy of a system and its surroundings is neither created nor destroyed. |
Second law | • In a spontaneous process, the total entropy of the system and surroundings increases: ΔSuniv > 0. For a reversible process, ΔSuniv = 0. |
E. Enthalpy change, ΔH
: ΔH = Hfinal – Hinitial
Exothermic processes, ΔH < 0 | Endothermic processes, ΔH > 0 | |
Diagrams | 
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Graph and equation | 
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System | • Giving off heat to surroundings • The system loses energy as heat; enthalpy decreases at constant pressure • Bond formation processes • State change: gas → liquid → solid | • Getting heat from surroundings • The system gains energy as heat; enthalpy increases at constant pressure • Bond breaking processes • State change: solid → liquid → gas |
Surroundings | • Surroundings receive heat | • Surroundings supply heat |
Examples | • Combustion of fuel, acid and base neutralization, catabolism of glucose in a cell, mixing water with strong acids | • Melting solid salts, producing glucose by photosynthesis, cold pack by mixing water with ammonium nitrate |
** The total energy of the system and its surroundings is conserved in both exothermic and endothermic processes.
F. Hess’s Law
- The ΔH of an overall reaction is equal to the sum of the ΔH of the individual steps.
- Enthalpy is a state function: its change depends only on the initial and final states, not on the pathway. Reversing a reaction changes the sign of ΔH; multiplying an equation by a factor multiplies ΔH by the same factor.
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- Question: Calculate the energy change, ΔH, for the reaction S(s) + O2(g) → SO2(g) from the following information.
For this calculation exercise, use the supplied hypothetical reaction enthalpies for the equations as written. These values are not tabulated standard formation enthalpies.
Table A | 
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Solution:
Table A | 
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1/2 × 1st reaction | 
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Reverse direction of 2nd reaction | 
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1st reaction + 2nd reaction | 
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Therefore, the energy change of the reaction S(s) + O2(g) → SO2(g) is +5 kJ/mol.
G. Measuring enthalpy change, ΔHo
: calculate reaction enthalpy from standard enthalpies of formation or estimate it from average bond energies. The standard enthalpy of formation, ΔH°f, refers to forming one mole of a substance from its elements in their reference states.
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Question: Using the supplied rounded formation enthalpies, calculate ΔHo for the reaction.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Molecules | 
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| -75 |
| -390 |
| -290 |
Solution: Using the supplied rounded values, ΔHo = Σ products of ΔHof – Σ reactants of ΔHof
= [CO2 + 2H2O] – [CH4] ** ΔHof of O2 = 0 kJ/mol
=[(-390) + 2(-290)] – (-75)
= −895 kJ for the reaction as written. Multiply each standard formation enthalpy by its stoichiometric coefficient; O2(g), an element in its reference state, has ΔH°f = 0.
Question: Estimate ΔH for the gas-phase reaction using the average bond dissociation energies in Table B.
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
Table B | |
Bonds | Average bond dissociation energies (kJ/mol) |
C-H | 410 |
C-C | 350 |
O=O | 500 |
C=O | 730 |
O-H | 460 |
Solution: Σ (Bond energies of reactants) – Σ (Bond energies of products)
= [(C-H)×12 + (C-C)×2 + (O=O)×7] – [(C=O)×8+(O-H)×12]
= [410×12 + 350×2 + 500×7] –[730×8+460×12]
= -2,240kJ
H. Entropy change, ΔS
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ΔS > 0 | ΔS < 0 |
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- ΔS = ∑ (entropy of products) – ∑ (entropy of reactants)
Question: Calculate ΔS for this hypothetical reaction using the supplied molar entropies in J/(mol·K).
W + X → Y + Z
S | |
W | 3 |
X | 2 |
Y | 5 |
Z | 4 |
ΔS = (Y + Z) – (W+X)
= (5+4) – (3 +2)
= 4 J/(mol·K)
- Entropy of the surroundings, ΔSsurr
Increasing entropy of surroundings | Decreasing entropy of surroundings |
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I. Gibbs free energy change, ΔG
: at constant temperature and pressure, ΔG determines whether a process is thermodynamically favorable in the forward direction. ΔG < 0 is favorable, ΔG = 0 indicates equilibrium, and ΔG > 0 favors the reverse direction. Thermodynamic favorability does not determine reaction speed.
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- ΔH < 0 and ΔS >0 at T > 0 is always a spontaneous reaction.
Question: For these hypothetical values, which choices represent a thermodynamically favorable reaction at T = 1 K? Use ΔH in kJ/mol and ΔS in kJ/(mol·K).
A. ΔH = -1, ΔS = -2
B. ΔH = +1, ΔS = -2
C. ΔH = +1, ΔS = +2
D. ΔH = -1, ΔS = +2
E. C and D
ANS: E
Solution: For ΔG = ΔH – TΔS
A. -1 – (-2) = +1
B. +1 – (-2) = +3
C. +1 – (+2) = -1
D. -1 – (+2) = -3
Question: For a hypothetical process with ΔH = 6 J/mol and ΔS = 3 J/(mol·K), calculate the equilibrium temperature in kelvin.
